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ESATIMATION: We represent two formulas for dividing the mean and the standard deviation of the binomial distribution: μ=np s=√npq where n= number of trails p= probability of success q=1-p=probability of failures Theoretically, the binomial distribution is the correct distribution is the distribution to use in constructing confidence intervals to estimate a population proportion. As the sample size increases, the binomial can be approximate by an appropriate normal distribution, which we can use to approximate the sampling distribution. Statisticians recommend that in estimation, n be the large enough for both np and nq to be at least 5 when n you can use the normal distribution as a substitute for the binomial. Let’s express the proportion of successes in a sample by p (p bar). Then modify equation 5-2,so that we can use it to drive the mean of the sampling distribution of the proportion of successes. In words, μ=np shows that the mean of the binomial distribution is equal to the product of number of trails, n, and the probability of successes; that is, np equals the mean number of successes. To change this number of successes to the proportion of successes, we divide np by n and get p alone. The mean in the left _hand side of the equation becomes μp, or the mean of sampling distribution of the proportion of successes. MEAN OF SAMPLING DISTRIBUTION OFR RHE PROPORTION μp=p…………………….[7-3] Similarly, we can modify the formula for the standard deviation of the binomial distribution, √npq, which measures the standard deviation in the number of successes ,we divide √npq by n and get √pq/n. In statistical terms, the standard deviation for the proportion of successes in a sample is symbolized and is called the standards error of the proportion. STANDARD ERROR OF THE PROPORTION Standard error of the proportion ……………….. [7-4] We can illustrate how to use these formulas if we estimate for a very organization what proportion of the employees prefer to provide their own retirement benefits in lieu of a company _sponsored plan. First, we conduct a simple random sample of 75 employees and find that 0.4 of them are interested in providing their own plans. Our results are n=75…..sample size p=0.4…..sample proportion Next, management requests that we can use this sample to find an interval about which they can be 99% confident that it contains the true population proportion. But what are p and q for the population? We can estimate the population parameters by substituting the corresponding sample statistics p and q (p bar and q bar) in the formula for the standard error of the proportion.*Doing this we get Now we can provide the estimate management needs by using the same procedure we have used previously. A 99% confidence level would include 49.5% of the area on either side of the mean in sampling distribution. The body of appendix Table 1 tells us that 0.495 of the area under the normal curve located b/w the mean and a point 2.58 standard errors from the mean. Thus, 99% of the area is contained b/w plus and minus 2.58 standard error from the mean. Our confidence limits then becomes P+2.58 Ồp =0.4+2.58(0.057) =0.4+0.147 =0.547……….upper confidence limit p-2.58 Ồp=0.42.58(0.057) =0.4 – 0.147 =0.253…………lower confidence limit thus, we estimate from our sample of 75 employees that with 99% confidence we believe that the proportion of the total population of employees who wish to their own retirement plans lies b/w 0.253 and 0.547.